Explanation Given data: The values of, the resistor R is 40 Ω , an inductor L is 2 H, a capacitor C is 0.0025 F, and battery E is 12 V. Write the expression for the differential equation, L Q ″ + R Q ′ + 1 C Q = E (1) Substitute 2 H for L , 40 Ω for R , 0.0025 F for C and 12 V for E in equation (1), 2 Q ″ + 40 Q ′ + 1 0.0025 Q = 12 2 Q ″ + 40 Q ′ + 400 Q = 12 (2) Consider the auxiliary equation, 2 r 2 + 40 r + 400 = 0 r 2 + 20 r + 200 = 0 (3) Roots of equation (3) are, r = − 20 ± ( 20 ) 2 − 4 ( 1 ) ( 200 ) 2 ( 1 ) { ∵ r = − b ± b 2 − 4 a c 2 a for the equation of a r 2 + b r + c = 0 } = − 20 ± i 20 2 = − 10 ± i 10 Write the expression for the complementary solution of two complex roots r = α ± i β , Q c ( t ) = e α t ( c 1 cos β t + c 2 sin β t ) (4) Substitute − 10 for α and 10 for β in equation (4), Q c ( t ) = e − 10 t ( c 1 cos 10 t + c 2 sin 10 t ) (5) The Right hand side (RHS) of a differential equation contains only a numerical value, and then the particular solution for this case can be expressed as follows

James Stewart Calculus for MAT 127/128/229 8th edition

8th Edition

ISBN: 9781305743663

Author: James Stewart

Publisher: Cengage Learning

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Chapter 17, Problem 19RE

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To find: The charge at time t.

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The charge at time t .

Solution Summary: The author explains that the charge at time t is underset_Q(t)=-0.02e-10t

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Chapter 17 Solutions